text:
Purpose of Reflex in Flying Wing Aircraft
I want to design a small RC flying wing aircraft and I am struggling to understand the purpose of reflexed airfoils.
From my understanding, for longitudinal stability, the purpose of the tail on a conventional aircraft is to shift the aerodynamic center of the vehicle sufficiently rearward such that the center of gravity can easily be placed in front of it. This allows for static stability to be achieved without significant nose weight.
With a combination of wing sweep and careful CG placement, it seems that, even without a tail, the aircraft could be made statically stable without reflex. I am struggling to understand, therefore, why it is so prevalent.
I have read and watched a lot of content online about the reflex generating a small downward force that creates a moment similarly to the tail, but I am not understanding why that torque is necessary if you already have a stable aircraft.
Any help would be greatly appreciated.

If the CG is in front of the AC then you don't need any stabilising tail, see for example hang gliders

A tail surface just aligned with the airflow to move the aerodynamic center back just creates a very stable ballistic missile or lawn dart. To fly, you have do things with the stabilizing surface to force the wing to operate at a positive angle to the flow.
You "trim" it with the horizontal tail, generally by creating a negative lift force that forces the body/wing to an angle that achieves a static equilibrium state (sort of weathervaning) while the wing is forced to a positive angle, where it makes lift, and along you go.
You end up with an opposing force balance, the wing's pitching moment and the C of Mass pulling the nose down, and the tail opposing it. Depending on how the tail surface is angled, you achieve that force balance at some positive angle to the flow, 2, 5 , 10 degrees, whatever to make lift to fly instead of being a lawn dart, and it will seek that angle when disturbed from it.
Static stability is the seeking of an angle of attack (you could think of it as weathervaning into that angle) and trim is the creation of an opposing pitch force balance point that controls what that sought angle of attack will be.
A reflexed flying wing does the exact same thing, but with a much shorter arm to work with, that is, the job of trimming the wing to seek a certain angle of attack by creating an opposing force balance, is moved from 10 feet back at the end of a fuselage to the trailing edge of the wing itself, the reflexed part. In other words, the reflex is just the tail's stabilizing surface, set at some angle to generate a trimming force, moved to the wing's trailing edge.
The same things are happening, but with the very short moment arm to work with, you end up with having to live with certain limitations imposed, mainly sensitivity in pitch, and a very narrow CoG range, because the trimming/maneuvering forces have so little leverage. And thus flying wings are not that common, except where the limitations can be accommodated.

For stability, ask yourself how forces and moments will change with a change in angle of attack. In other words: Do not look at the force equilibrium, but at its derivative with respect to angle of attack. Or speed.
It is easy enough to bring the center of gravity close to the center of pressure - fine tuning is then done with the elevator. But the important part is that the airplane will do that fine tuning itself once the equilibrium is distorted.
The function of a tail surface is not to add some area at the rear - instead, it must be designed to change lift more than the wing, relative to their starting position, when something changes (this something being speed or angle of attack, mostly). How that is done is explained here.
Tailless designs still need something that provides this relatively larger lift change at the rear in order to be stable. This is done by adding a lightly or even negatively loaded area at the rear of the airfoil. Since the starting condition is a light load, any change in angle of attack will produce a larger change in lift relative to the equilibrium condition there than on the forward wing section.
With sweep, a second option opens up which can be used in parallel or alternatively to a reflex airfoil: Washout. By giving the outer wing a much smaller angle of attack, again the rear part of the wing is loaded less, making any change relatively stronger there than in the more loaded, inner section of the wing.


text:
Best way to mentally calculate wind correction angle (WCA) when in a holding pattern?
I have been out of flying for a while and need to brush up on my holding patterns. I need a quick method or rule of thumb to calculate the wind correction angle (WCA) for example when flying outbound for a holding pattern.
Can anyone provide a quick mental method or rule of thumb?
Thanks

I understand the formulas but this is not practical during flight

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Unmanned aerial vehicles communication link: frequency/spectrum?
I am studying airborne platform:

HAP - high altitude platform
LAP- low altitude platform
UAV - unmanned aerial vehicle

ITU gives frequency allocation bands for link between ground and HAP/LAP/UAV, but there is no information what the frequency is used in the link between them: HAP to HAP link, LAP to LAP link, UAV to UAV link.
I have heard that it is possible to use the same frequency as it was allocated for the inter-satellite link, but I am unsure if it is correct.
Recently I have found a paper "On the impact of inter-UAV communications
interference in the 2.4 GHz band" ( sorry didn't find it in open access). we study the performance of communication links between drones in the 2.4 GHz wireless band. It is WiFi 2.4 GHz band.
Does anyone know any regulatory documents for inter-platform links?

What about UAV-LAP links, and manned aircraft-UAV|LAP|HAP communication? Is that inside the scope of your question as well?

@DeltaLima, yes, it is. Honestly, I am confused about LAP and UAV. In one literature, UAV is LAP platform. So I have decided to assume LAP and UAV as two different platforms

UAV is a generic term for any unmanned flying thingy. LAP and HAP are both platforms: this term is typically used for a vehicle that is carrying either sensors (typically earth observation, but could also be atmospheric measurements) or communication infrastructure (networking). When we are talking about using a UAV for transportation of goods, the terms HAP or LAP are not used.

for inter UAV communication, there are several use-cases, and each may use a different link frequency. For example, inter aircraft communication for traffic management (traffic separation/detect & avoid, collision avoidance) will likely use a different frequency than payload data communication. Yet another frequency may be used for command and control purposes. Are you looking for any use-case specifically?

@DeltaLima. I did not consider any use cases yet. Currently, I would like to define frequency allocation for inter-platform links or satellite-platform link. I have read it is considered an optical link between HAP and satellite ( LEO or GEO) and HAP-HAP link, but optical between drones, I am unsure if it is possible.

@DeltaLima Is there a document with frequency allocation/spectrum for inter-platform communication? I read it will be under discussion in WRC2023. ok, but we have already had a lot of UAV devices in the air communicating with each other. They have already used a spectrum.

@DeltaLima As I know, please correct me if I am wrong, drones use today's wifi for communication with each other.  In one another forum, an user suggest me to use any frequency allocated for inter-satellite link except aeronautical frequencies.

@DeltaLima where can I find the spectrum for use cases you mentioned in one of the your comments (traffic management and payload data communication)?

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What is this aircraft seen at Peachtree City, GA?

Sorry for the poor quality, but this jet was taking off from Peachtree City Regional on Sunday 13 May. Anyone know the aircraft?


text:
How will the Rolls Royce UltraFan provide reverse thrust?
When perusing the Wikipedia article on the upcoming Rolls Royce UltraFan engine, I noticed that it cited an interesting bit of information from an Aviation Week article:

As the fan system is designed to vary pitch in all phases of flight, including landing, the nacelle will not include a thrust reverser.

I understand that a turboprop can enter the beta range in order to vary the blade pitch into the negative range in order to supply reverse thrust.
However, my understanding of the UltraFan was that it is to be a geared turbofan design with the fan and the core ducted inside a more-or-less conventional turbofan nacelle (minus the thrust reverser, apparently.)
My question, then, is how does the airflow work through this engine with negative fan blade pitch?
When the blade pitch goes negative, then the fan will be blowing air out the intake rather than sucking it in through the intake, no? Where then is the air coming from which is getting pushed out the front and where is the air coming from that is still needing to go in the normal flow direction through the core? Does the flow direction through the bypass ducts reverse?

The Wikipedia article further says "Along with eliminating the thrust reverser, a short slim nacelle would be lighter and less draggy, but in reverse-thrust the flow would be distorted, having to be turned around the nozzle into the bypass duct, and then partly reversed again into the intermediate compressor.", but unfortunately they do not give a source for this.

The flow direction reverses only through the bypass duct pulling by blades with negative pitch. A portion of it will still be sucked by low pressure compressor making 180° turn into the core. Having to counteract the losses in turn will obviously reduce the performance.
As stated in research paper by the team funded by Rolls-Royce plc: "the core mass flow rate was 57% of that at take-off conditions, for bypass duct flow rates of 74% of total take-off mass flow."
Another design by Pratt & Whitney features reversed position of the whole core requiring two turns of the flow for incoming and outgoing air. So, these turns are not a big problem as we see them traded for other pros.


text:
What way does induced downwash make a wing roll?
Say you have a wing, no sweep or anything to simplify things. One side of it, let's just say the left, magically doesn't have a tip vortex, therefore no induced downwash from the tip vortex. Would the wing (assuming the left side was still was making lift), roll to the right?
I assumed it would because the induced downwash on the right side of the wing would 'pull' the right wingtip down. Is that correct?

Anyone know why the downvote? I don't think this is a direct duplicate or anything

Related: https://aviation.stackexchange.com/q/81859/34686

Related: https://aviation.stackexchange.com/q/50301/34686

Downwash is not a force.  It is a small increment in the velocity vector.
Downwash can change the local angle of attack -- which can change the lift -- which is a force.
However, A wing has a rolling moment because the force on the two sides is not equal.  Period.  Full Stop.  If the lift force on the left wing is greater than on the right, the wing will roll to the right.  It is that smple.
We can talk about where the force imbalance comes from -- but you won't have a moment without a force imbalance.

Ah okay, I think I see what you mean. The downwash changes the local AoA and therefore the wing with the induced downwash would be the down going wing in the roll? I guess I'm basically asking if the wing side with the tip vortex would still roll downward, knowing this logic you provided.

Lets start with the time when the wing is not rolling, but with one wake magically eliminated.  The wing that sees greater downwash effects (the right wing) would see a decrease in the local angle of attack.  It would therefore make less lift.  There would be a rolling moment to the right -- which would initiate a roll.

Ah, that's what I was thinking, thanks!

Isn't the downwash (the flow of air deflected down) induced by the airfoil profile, as a consequence of air acceleration by the top side of the airfoil?

@mins We often model a wing as a horse shoe vortex.  One bound vortex at the quarter-chord of the wing.  Two trailing vortices representing the wake.  All of these  contribute to downwash.  The bound vortex is what you're talking about -- it is equivalent to replacing a 2D airfoil with a vortex at the quarter chord.  In 3D, the trailing vortices also contribute to the downwash -- they are what cause the lift to go to zero at wingtips, they are the driver of 3D effects.


text:
Why are some Chinese air to air missiles blue?

Why are some Chinese missiles blue? What purpose does this serve? I couldn’t seem to find any research on this.

Possibly those are inert "missiles". US Air Force and Navy do similar.
Here, an AIM-9 on F-16 wingtip launcher.

(pic from af.mil)

Ah okay, yeah that makes sense, thanks!

Confirmed.


text:
What happens to a shockwave when it goes through different temp/density air?
Specifically I am asking what happens to a shock as it travels through different density air. This answer's comments has some good information about it.
As said in the linked answer, a shock will bend the streamline next to it and extend the shock in doing so, which will also make the streamline next to it bend, which will also continue the shock, then etc etc etc. (read linked answer for a more comprehensive explanation) So knowing that, what does different density air do to this process?
My thinking was that it’d make the streamline trying to bend have some compressibility (higher temp = more compressible air), which would make it so the streamline bent less, which would change the flow turning angle, which would also change the shock angle. That doesn’t really make sense  as all the pictures of shocks I’ve seen are straight, even if they traveled through a different temperature/density air.
Something that would answer my question pretty easily: When the shock travels through a different temperature air, does that also make the shock bend?
The closest thing I’ve found on this subject is this Wikipedia article on atmospheric focusing, which touches on this topic slightly.

Higher temperature does not make the air more compressible.
Think of the air in layers.
Assume zero wind -- all the layers of air are still.  The thing that is moving (through still air) is the object -- say at 1700 ft/s.  Furthermore, the object causes the flow to turn 10 degrees.
At each layer, those things stay the same -- $V=1700$ ft/s speed, 10 deg turning.
So, for each layer, find the temperature.  From that, calculate the speed of sound.
$a=\sqrt{\gamma R\, T}$
Then the Mach number.
$M=V/a$
Then look up your shock chart, or use your shock formulas.  You know M and the turning angle $\theta$, so find the shock angle $\beta$.

Oh okay. So isn’t it easier to compress less dense air compared to denser air. (Here I’m assuming less dense air is higher temperature)

Compressibility from a fluid dynamic standpoint isn't exactly the same thing as putting gas in a bottle with a piston and pushing on the piston.  When we talk about a subsonic flow being incompressible vs. a high speed flow being compressible, it isn't an easy/hard thing, it is about how fast the molecules can react to what is happening around it vs. how fast those things are happening.  The ratio of the speed of things happening (Velocity) compared to the speed of molecules reacting (speed of sound) is the Mach number.

So in a very real sense, the Mach number is a measure of fluid dynamic compressibility.
The speed of sound increases with temperature $a=\sqrt{\gamma R T}$, but from that sense, for the same velocity, hotter air will have a higher speed of sound, a lower Mach number, and will experience less effects from compressibility for an otherwise similar flow.

oh okay, so even though it is more compressible in the sense easier to 'squeeze', the speed of sound is higher so the flow is still able to turn to continue the shock?

I still don't know that it is easier to squeeze.  If the two bottles of gas are at the same pressure, but different temperatures, then 'pushing' against them would take the same effort (force) and doing so through a distance (say to halve the volume) would take the same amount of work (energy) (work is force times distance).  So I would say that higher temperature does not make a gas easier or harder to squeeze.

I see, but in the atmosphere there isn't anything holding the gas from expanding therefore making it more compressible (at higher temperatures), right?

The fact that there is nothing containing the gas is exactly why compressibility is measured by the ratio of speed to the speed of sound -- compressibility is the ratio of how fast the flow can react to how fast changes are being forced on the flow.

When we're talking about changes of temperature through the atmosphere, we're talking about getting colder by 3.5 deg F per 1000 ft altitude increase.  So a very slow change in temperature.

Oh okay, thanks. So let me make sure I have this right : The higher temperature will affect the compressibility, (ratio of speed of sound to the speed of disturbance) therefore  affecting turning angle (angle of shock) and the strength of the shock?

Yes, but it is a small effect.


text:
Why is the upper surface shock further forward than the bottom shock on this airfoil?

If you look at this picture, you’ll see the shock on the upper surface is further forward on the wing, compared to the lower shock. Why is that?
(This is at Mach 0.9, as shown at the bottom of the image)

Because air has accelerated faster on the upper surface than the lower one?

As Sophit says -- the air has accelerated more over the upper surface, so the shock happens sooner.

Ah okay, thanks. Just to refresh my memory, don’t the upper surface-normal shocks happen when the pressure recovery starts on the airfoil?

There is a region of supersonic flow in front of the shock.  You can't see the forward boundary of this region.  There is some back pressure associated with the pressure recovery.  The location of the shock depends on how high the Mach number gets in the supersonic region and also the strength of the back pressure.  It is a complex balance that works out differently on the upper vs. lower surface because this airfoil is at angle of attack (and/or is not symmetrical) such that it is producing lift.

Oh okay, thanks. It makes sense that the bottom surface of an airfoil would have a shock, because the first front 1/4 of the bottom surface makes the same amount of lift as the last 3/4 of the lower surface. This makes the same effect as the upper surface as the high pressure goes to the low pressure (Bernoulli). (Got that from the comments of this answer)


text:
Can anyone recommend me some aviation textbooks?
I don't go to school for aviation, just a hobby/interest of mine. Would like to learn about things like vortices, flow patterns, pitch stability, effects of wing planform, and any other related stuff. Preferably around 100$ give or take, from the prices that I've seen. Thanks!

Introduction to flight by John Anderson, fluent, relatively easy to read, rigorous and correct, a must have. I'd leave pre- and post-war books alone, lot has changed since then.

Why not start with a private pilot training manual?

In the old days, all of these topics would be in the same book.  Most modern books (even at the undergrad level) are pretty specialized.  I.e. an aerodynamics book won't get into stability and control or aircraft performance.
Aerodynamics for Naval Aviators by H. H. Hurt Jr..  Available as a free download from the FAA.  Not super rigorous (i.e. not full of equations and derivations).  It is the standard for teaching pilots just enough about the airplane that they can get by.  Also available as an inexpensive paperback if you like to read paper books.
Engineering Analysis of Flight Vehicles by Holt Ashley  Available as an inexpensive Dover book.  One of the more modern books that cover the breadth of topics you're asking for.
Theory of Flight by von Mises  One of the old classics.  A good example of what aeronautics books were like pre-war.  Also an inexpensive Dover book.
Airplane Performance, Stability, and Control by Perkins and Hage  This one is from 1949 and is not readily available today.  However, it pretty much set the standard for modern performance and S&C books.  Most current books owe a lot to Perkins and Hage.  Keep an eye out and buy a cheap used copy on eBay if you can.

Oh okay I’ll try some of those out, thanks!


text:
What will the pressure of accelerated air be relative to the pressure that accelerated it?
Imagine you have a high pressure reservoir filled with air. This reservoir is 1 psi
above ambient atmospheric pressure. There is also a valve attached to the reservoir.
Now, open this valve. The air will rush and accelerate out, and due to Bernoulli's principle, this fast air will be low-pressure.
Relative to the atmospheric pressure, what will the pressure of the accelerated air be at its maximum velocity? Will the low-pressure-peak be 1 psi below ambient? 0.5 psi?

This applies to aviation when the air travels to the front of an airfoil, accelerates over the top and becomes low pressure. I'm wondering the ratio between the pressure that accelerated the air over the foil vs. the pressure of the accelerated air itself.
I used the analogy of the pressure reservoir because I couldn't find a way to elegantly explain it using an airfoil, but the same principle still applies.

If there is anything that I missed, please comment and I will happily edit it in (or out)!

This belongs more to physics than aviation

Did we learn from jets that max velocity is at ambient pressure?  Certainly not less (that would slow the flow) (air rushing out of a high pressure container will have higher than ambient pressure)

Don't send aerodynamics questions over to Physics, it'll never get answered properly.

@RobMcDonald I actually had asked this on Physics.SE, and it never got an answer after multiple days, so I asked it here.

@RobMcDonald: for how much I like this kind of question and your perfect answer sorry but this whole story has nothing to do with airplanes. Otherwise I could ask any question about bikes because they have wheels just like airplanes...

Let's call the valve a nozzle.
In the reservoir, we have:
$P=P_{res}=P_{amb}+1\, \mathrm{psi}$
$V=V_{res}=0$
$\rho=\rho_{res}=\frac{P_{res}}{R\,T_{res}}$
The temperature is unspecified
$T=T_{res}$
If we assume a non-insulated reservoir that has been sitting for long time, then we can assume $T_{res}=T_{amb}$.
Next, let us assume that the nozzle is large enough that we can ignore viscous effects.  Let us also assume the pressure difference (1 psi) is small enough that we don't get into substantial compressible effects.
With these assumptions, Bernoulli's equation is all we need.  First, calculate the total pressure (stagnation pressure) inside the reservoir...
$P_0=P+0.5\,\rho\,V^2$
$P_{0,res}=P_{res}+0.5\,\rho_{res}\,(0)^2$
I.e. the total pressure is the pressure -- precisely because the flow is stagnant.
$P_{0,res}=P_{res}$
This flow will pass through the orifice without viscous effects, work addition (or removal), passing through shocks, etc.  We're also assuming incompressible for now.
These conditions are sufficient for the stagnation pressure to be a constant.  When the flow expands outside the reservoir, it will have the exact same total pressure.  However, it will expand to match the ambient pressure $P=P_{amb}$.  Since it is incompressible, the density will be constant.
Total pressure is constant
$P_{0,jet}=P_{0,res}=P_{res}$
Jet expands to ambient pressure
$P_{jet}=P_{amb}$
Jet is incompressible
$\rho_{jet}=\rho_{res}$
Write Bernoulli's equation for the jet...
$P_{0,jet}=P_{jet}+0.5\,\rho_{jet}\,{V_{jet}}^2$
Solve for $V_{jet}$
$V_{jet}=\sqrt{ \frac{P_{0,jet}-P_{jet}}{0.5\,\rho_{jet}}  }$
Because total pressure is constant and the jet expands to ambient pressure, we know the pressure difference in the numerator of the radical is 1 psi.
$V_{jet}=\sqrt{ \frac{1\,\mathrm{psi}}{0.5\,\rho_{jet}}}$
Substituting a few things we know
$V_{jet}=\sqrt{\frac{2\times 1\,\mathrm{psi}\,R\,T_{amb}}{P_{amb}+1\,\mathrm{psi}}}$
And converting psi to psf (pounds per square foot) because our density will be in slugs per cubic foot and Temperature in Rankine, etc.
$V_{jet}=\sqrt{\frac{288\,\mathrm{psf}\,R\,T_{amb}}{P_{amb}+144\,\mathrm{psf}}}$
Standard conditions.
$P_{amb}=2116.22 psf$
$T_{amb}=518.67 deg\, R$
$R=1716.46 ft\ lbf / (slug\ deg\,R)$
Doing the math
$V_{jet}=336.8 \,ft/sec$
Which is pretty reasonably for it still being incompressible.
A higher pressure ratio would violate incompressibility and you would not be able to use Bernoulli's equation.  Instead, in that case, you would want to use the Isentropic Flow Relations.  These are a set of equations that apply in a similar way, but that work for compressible flow.
Importantly, the isentropic flow relations calculate the ratio of total or stagnation properties to the ambient properties dependent on the Mach number.
Since you are starting from a reservoir, you would use the fact that velocity and Mach are zero in the reservoir and assign the reservoir properties as the total or stagnation properties.
Then, you can use the isentropic flow relations to expand the flow out to ambient.  We start with the assumption that the flow is perfectly expanded.  In that case, we use the Pressure equation of the isentropic flow relations to find the Mach number in the jet.
I'm pretty sure that if that Mach number is subsonic, then we're done.
If that Mach number is supersonic, then we need to understand the geometry of the nozzle more -- specifically, we need to know the area of the throat and the area of the exit.
We then treat it as a converging-diverging nozzle, with choked flow at the throat and expand the flow out to either an under-expanded or over-expanded condition at the nozzle exit.
I might have some details around how we determine if the nozzle flow is subsonic or supersonic, but you get the general idea.

Thanks for the very detailed answer! So where it says P= T= R=, are those the values required to get the psi (or psf)? If my inference is correct, 336.8 ft/s is the velocity that the air will be leaving the reservoir (or nozzle) at.

yes, if you plug in the three values (P,T,R) listed under standard conditions into the equation for Vjet above, you get the value of 336.8 ft/s given.  I will modify my answer to clarify those values a bit.

I see, thanks. So lastly, is there a way to convert 336 ft/s to the actual pressure of the accelerated stream? (Like 0.5 psi, or psf if it’s easier of the atmosphere pressure?) Sorry if you’ve already answered this, and I’ve not followed.

The pressure in the stream is ambient pressure.  The pressure is higher in the reservoir, it expands to ambient.

Very good and informative answer but... where's the airplane??

As I've pointed out before, there is no Aerospace Engineering forum on SE, so this one ends up fielding many aerodynamics questions.  You can't have a more fundamental aerodynamics question than an elementary homework problem applying Bernoulli's equation.  Of course it is also how both jet and rocket engine nozzles work.

@RobMcDonald Oh okay, thanks for pointing that out. So, last follow up question, but if you were to install an array of pressure sensors along the whole jet, how low would the lowest pressure-peak be? (As in psi or psf) Again, sorry if you’ve already answered this, and I’ve not followed.

We can't answer that using this level of theory.  At this level of abstraction, the flow is either 1) Stagnant in the reservoir or 2) in the jet.  Using Bernoulli, we could know the velocity that corresponds to any pressure between Pres and Pamb.  However, we don't have any spatial / geometric idea of how and where the flow accelerates.  Obviously, the flow must move somewhere inside the reservoir -- or else nothing would exit the orifice.  Likewise, we pay no attention to the shape of the orifice, if it is a tube, how long it is, etc.

@RobMcDonald one minor adjustment in the calcs may be that the air would cool from expansion, but would also heat exchange with the ambient air. It would probably depend on the nozzle as to what T would be when P of plume = P ambient.

@RobertDiGiovanni This calculation accounts for Tjet != Tamb, I just didn't bother making the calculation of what Tjet would be. For simplicity, I rearranged things to be in terms of quantities known up-front. If I was doing this in practice, I would have used more intermediate calculations that would have given values to other quantities along the way -- for example, rho_res=rho_jet is never calculated either. The temperature effects you're after will become much more significant as you deal with higher pressure ratios and the compressible case that necessitates the isentropic flow relations.


text:
How wide would an A330's wingtip vortices be around a minute after landing?
I recently watched MenTour Pilot's new video about the Royal Air Maroc 737 takeoff from Frankfurt several years ago that made the news at the time because it rotated, failed to fly away at high AoA for several seconds, lowered pitch and settled back to the runway, accelerated some more, and then took off normally.
The original video of the incident is here:

In his video, MenTour Pilot suggests that the 737 was prevented from taking off because of the sudden drop in airspeed from encountering the left wingtip vortex of the A330 that had recently landed on a nearly perpendicular runway. However, the original video seems to show the RAM 737 at high AoA for around 11 seconds after its first rotation attempt. At Vr for a 737, that seems like it should burn through 2,000 to 3,000 feet of runway, depending on weight. (135 kt = 228 ft/s, for example)
So, my question is, how wide is each wingtip vortex of a landing A330 around 1 minute after landing? The conditions appear to be almost calm wind. Would each vortex really be thousands of feet wide by that point?

Without knowing the details of this specific incident, it has happened before that a too low take-off mass has been used for performance calculations. The resulting Vr will be lower and the situation may be aggravated by choosing a lower power setting (derate / assumed temperature) to save on engi e maintenance. In that case most of the runway will be used by the time the crew discovers the aircraft does not fly at Vr

@DeltaLima Yep. And, personally, I think that's probably what happened (computed either for wrong mass, a different t/o config, or both,) but I was curious how big the wingtip vortices would actually be at that point anyway. Thousands of feet seemed really excessive. I know when I've hit my own wake in light aircraft, even after making a full 360 turn, the spread was almost nil and hitting the wake was more just like a single jolt, not an effect spread over time. But, of course, an A330 is much larger and faster than a PA-28. - haha

During last summer I've spent a few days time  at & near the taxiway next to position where this aircraft got airborne for the first time. I don't remember seeing any aircraft rotating that early, and due to lack of passengers, take-off weights were low this summer. Another hint is that rotation is approximately 20 seconds after application of take-off thrust. In my experience, the rotation is usually after ~33 seconds on the 737-800, perhaps a few seconds earlier on the -700. This was a very early rotation by all means.


text:
Why does the airflow turn less when you extend further from the body?
In this answer, it says as follows:

Because "far enough" from the Concorde's nose, the path followed by the airflow is not affected by its presence and it's just a straight line. Close to the nose we need a shock wave at 45° to make the airflow turn the 15° of the nose. Anyway, a bit further, the flow turns a bit less than 15° (let's say 10°) and therefore a shock wave of 40° is enough to achieve this inclination

Why is the flow allowed to turn less further from the body, or the Concorde in this case? What changes further from the body to allow the flow to turn less?
I don’t really understand why

"far enough" from the Concorde's nose, the path followed by the airflow is not affected by its presence and it's just a straight line

This fact would change the flow turning angle. The shock needs to turn a set angle unless it is changed.
The explanation might be talking about the 3D relieving effect, but I don’t think so because that would make the explanation false for 2D objects.

Imagine the plane "flies" through a dense stack of flexible pipes, filled with air. Normally, the pipes would fill up all of the available volume. So how would the additional volume of the airplane fit in?
First, turn things around to simplify them: Imagine that the air in the pipes is moving at airspeed while the aircraft is stationary. Now you take the position of an observer moving with the airplane. The physics is the same, only your point of view has changed.
The answer is: By reducing the diameter of the pipes locally in the vicinity of the airplane! In subsonic flight this happens by acceleration of the air, so the air in the pipes around the plane flows faster, taking up less space. Once the airplane has passed (to be precise, when its cross section narrows down towards the tail), the air in the pipes assumes its old speed and the pipe cross section will increase back to its initial value.
In supersonic flight, things turn around because now the change in density dominates the pipe cross section: Compression shocks will not only slow the air down, but also compress it so much that the pipe cross section shrinks.  That shrinkage might require that several pipes away there is still some shrinkage required, but if the air in enough pipes is compressed, enough volume has been freed up for the airplane to fit in. The pipes farther away do not need to shrink and can stay at their initial density. In other words, the strength of the shock has died down to a value where no further density change is necessary anymore.
In reality, the density change will die down with the square of the distance from the airplane, but never disappear completely. The presence of the airplane does affect all the pipes, but this effect will be so small at a distance that we can assume as a good approximation that the presence of the airplane is not felt anymore.

I see. So basically the density changes will change the pipe(s) cross section enough around the aircraft which will also change the angle (and also therefore strength) of the shockwave? (and when the presence of the plane isn't felt anymore, this changing of shockwave angle will stop)

@Wyatt The shock angle at a distance is determined by the Mach number. Close to the airplane it is a combination of Mach number and contour. For air to speed up and density to increase you need an expansion fan.

"The shock angle at a distance is determined by the Mach number. Close to the airplane it is a combination of Mach number and contour." That's exactly what I've actually been trying to figure out (figure out why this that is the case), but I couldn't effectively explain it. (I wonder if that would require a separate question)

@Wyatt How far will the airplane have moved when the pressure change it produces reaches a point in the surrounding? Just think how far that pressure change can travel in the same time the plane travels. Make a sketch of it and look at different timesteps. That should help in understanding.

I think I see where you’re coming from. The pressure variation can’t travel faster than the speed of sound away from the aircraft, so it’s forced to be at a certain angle. Thanks for helping me figure this out!

Just to really make sure I have this correct, because in the past I have marked the question as 'solved' in my head when I was incorrect: According to the oblique shock chart, at Mach 2 the Mach cone will be 30°. A deflection of 5° will make a shock of ~34°. This 34° would be greater than the 30° of the Mach cone. But because of the effect you mentioned about how fast the disturbance can travel outwards, it would force the 34° shock to be 30°, AKA the same angle as the Mach cone. Is this correct?

@Wyatt Of course not. The air is deflected by 5°, so it does not flow in the same direction as before. With that new direction, the Mach angle will become wider when seen from the outside. When you place yourself as an observer inside that new flow direction, the Mach angle will still be 30°.

Just to firstly make sure I have the numbers right, in the end with the scenario I described the Mach cone would be 34° and also the shockwave? (I’m thinking the Mach cone would be 34° because if it is still 30° relative to the new already turned flow, then it’d be at a greater value relative to the original flow) That might not make sense, understandably, but I believe I could explain it in a dedicated comment.

@Wyatt The oncoming flow is irrelevant for the shock cone. It is the flow past the shock and its flow direction which is relevant. And a Mach cone is defined by its shockwave. They are the same thing, essentially.

Ah, I think I now understand what you mean. Again, thanks a lot. (Sorry for the late reply!)


text:
Does the pressure recovery on an airfoil cause separation more with turbulent BL’s?
If you have a turbulent boundary layer going through pressure recovery on an airfoil, why is it more likely to separate than a laminar boundary layer?
From my research/intuition I think turbulent boundary layers are less likely to separate because they have more mixing of flow.
What sparked this question is in this answer, it talks about “boundary layer tripping” and I found here that means where you make flow turbulent on purpose. Also in that answer it talked about how tripping the boundary layer made separation happen easy.
So basically it said that tripping the BL will cause separation to happen easier because of the pressure recovery not being handled as well and I’m wondering why a turbulent BL wouldn’t handle pressure recovery as well as a laminar BL.

You have this backwards.  A turbulent boundary layer is more likely to stay attached.
The second article you link to indicates that turbulent BL's resist separation (correct).
The first article you link is dealing with a very specific case of a laminar airfoil -- some laminar airfoils have different behavior that is more complex than just the boundary layer's ability to penetrate an adverse pressure gradient.
The intentional boundary layer tripping in the second article is not usually applied to flight vehicles.  It is usually applied to wind tunnel models where you want to simulate the full-scale turbulent behavior, but your model scale would naturally result in laminar flow.
Edit:
An example pressure distribution for a laminar flow airfoil is provided below.

The 'Laminar Rooftop' is a region of the pressure distribution where  there is an extended proverse (sloped up) or neutral (flat) pressure gradient.  This is different from conventional airfoils that typically have a distinct pressure peak near the leading edge and an adverse (sloped down) gradient for the remainder.
Laminar flow can have separation bubbles -- where flow separates from some forward part of the airfoil and then re-attaches slightly later.
Turbulent flow typically separates from the trailing edge first -- as angle of attack is increased, the separation point moves forward.
Consequently, typical turbulent separation is gentle -- at first just a small amount of your airfoil has separated flow.  This increases gradually as you increase alpha.  This results in a $C_L$ vs. $\alpha$ stall curve that is gentle and continues to produce a good amount of lift over a large range.
Conversely, some laminar flow airfoils have much more abrupt separation.  They hold their lift for a good while -- but then they just let go and abruptly lose everything -- because they stall from the leading edge first).

Ah okay, thanks. So how might a laminar airfoil have more complex behavior than a normal one? Aren't laminar airfoils just airfoils that try to avoid strong pressure gradients?

After doing some more research, I found this answer in which it talks about this subject. So from what I read from that, laminar airfoils will be more susceptible to separation because they're laminar, and you need to transition the flow to turbulent for it to stay attached. Is this (at least part) of what is meant by different behavior?


text:
Is a stream of air / jet exhaust low pressure?
Inspired by this answer, is a moving stream of air (Jet exhaust, rocket plume, etc) low pressure?
I’ve always thought that a higher pressure area (say the combustion chamber of a rocket engine) will be high pressure, then the lower pressure atmospheric air will ‘pull’ on the higher pressure combustion chamber gas therefore kind of stretching it out. (I know air never really pulls anything, but it helps to think of it that way)
I found this when doing my own research and in it it says :

When this low-pressure, supersonic jet leaves the nozzle and is confronted with ambient air, the air pressure will compact the jet.

So it seems like this would confirm that a moving jet of air is low pressure, right? Am I missing something?

Do you want to know for subsonic (EPR value) or supersonic velocities? It seems you are mentioning only supersonic engines.

@mins yeah good point. Shockwaves do complicate things, so for simplicity I'll say just below supersonic values / fairly subsonic values.

The pressure in a stream of gas might be above or below ambient pressure.
In the context of the quote, the purpose of a bell aka skirt aka nozzle in a rocket engine is to convert low speed high pressure gas into high speed gas, to increase thrust.  That conversion happens to also reduce the pressure, per Bernoulli.  That reduction might go below ambient, or not.  The one generalization you can make is that the pressure in the exhaust stream is lower than that at the nozzle throat.

Ah okay. So won't the atmosphere do the job of the convergent-divergent nozzle? Not as efficiently maybe, but won't the atmosphere do the same thing?

The "nozzle" converts pressure into speed.  mv of the exhaust is thrust.  Ideally, P at nozzle exit = ambient pressure.  The plume (theoretically) contracts as it cools, but the exchange of exhaust energy to the surrounding air is all after the fact of propulsion.  As long as nozzle pressure is higher, surrounding air pressure cannot do the job of steel.

@RobertDiGiovanni Oh okay, but won't the higher pressure exhaust gas expand, like it  does with a C-D nozzle? (pretend there are 2 cases, one without the C-D nozzle and one with)

You really don't want it to expand, but it's hard to be perfect.  I know standing behind a jet is a bad idea.  Essentially, you have a torrent of high speed air (that entrains ambient air). Your question about the pressure of the stream is interesting, perhaps with pressure as potential energy and the moving stream as kinetic energy.

@RobertDiGiovanni Yeah good points. My guess is that the lower pressure outside atmosphere air will 'stretch' out the higher pressure gas inside the jet engine / rocket engine. I know air doesn't pull anything in that sense, but it helps to think of it that way, as the same effect still happens. That still doesn't mean the stream will be low pressure, as the stretching effect might not be powerful enough at such high pressure (though downstream it might catch up)


text:
Why do the shocks on the upper and lower surfaces of a wing stop at the trailing edge?
Why do the shocks on the upper and lower surfaces of a wing stop at the trailing edge? When going Mach 0.82, the shocks on the upper (and lower) surface don't go the whole chord length of the wing. (In this picture, the upper shock stops around half way)
When going around Mach 0.95, the shocks make it all the way to the trailing edge. I was told that if you go faster, the shocks won't move back more past the trailing edge, like they would if you were speeding up from Mach 0.82. Why is that?

I think that @RobMcDonald has already answered you this question

@sophit Oh, would you mind telling me what part of that answer answers my question? I can't seem to recognize it.

"When flow regions 1 and 3 come together, they can't continue in a straight line. They must turn. That turning is accomplished by a shock." and "So, you can see that the shock 'stops' at the trailing edge because it must be there to turn the flow. After the TE, there is no need to turn the flow." are the relevant parts.

The shock on the surface of the airfoil is a normal shock.  It exists because the flow is supersonic, but is facing a back pressure such that it must shock down to subsonic.  The flow does not turn at that shock because the surface is locally smooth.
Zoom in on the TE of the airfoil...

Flow regions 1 and 3 are supersonic.  2 and 4 may or may not be -- depending on the velocities and angles involved.
When flow regions 1 and 3 come together, they can't continue in a straight line.  They must turn.  That turning is accomplished by a shock.
The dotted line between regions 2 and 4 is called a slip line.  Velocity across the slip line must be parallel (but not necessarily equal in magnitude).  Likewise, pressure across the slip line must be equal.  You can think of the slip line as a thin flexible membrane -- flow must be parallel to the slip line and if pressure was higher on one side, it would push the membrane until pressure was equal again.
So, you can see that the shock 'stops' at the trailing edge because it must be there to turn the flow.  After the TE, there is no need to turn the flow.
There are situations where there can be a third shock in regions 2 & 4, but that is actually a small normal shock that serves to slow the flow down to subsonic.
EDIT:  Adding some more details inspired by followup questions...
Even a thin boundary layer can complicate matters dramatically.
The shock causes an abrupt rise in pressure.  This is an 'adverse' pressure gradient.  Boundary layers tend to separate in adverse pressure gradients, so we often get separation at the point of the shock.
This separation bubble 'looks like' a bump in the surface to the flow outside the boundary layer.  So you now need some turning where before you didn't.  This causes the shock to change shape.
Shock-boundary layer interaction is really complex.  You can see evidence of shock-boundary layer interaction in the Mach 0.84 to 0.98 images below.

Here is an extensive paper on the subject.  Check out figures 9, 10, 11 in the paper.  They focus specifically on the difference between an idealized inviscid transonic flow and a transonic flow with a boundary layer.
This paper calls part of this phenomena a delta shock (referring specifically to the compression waves at the front of the boundary layer interaction.  I am less familiar with that name -- I usually see this whole arrangement (slightly bigger scale) referred to as a lambda-shock because it looks like the greek letter lambda $\lambda$.
The cartoon I drew before was an inviscid idealization.  You can see how a little bit of boundary layer can significantly complicate matters.
Here is another idealized cartoon, but zoomed way in, 'just before' the shocks reach the trailing edge...

Here the upper and lower surface shocks are normal shocks, they meet the surface at 90 degrees and do not turn the flow.  Also, the regions downstream of them (2 and 4) are now subsonic flows.
Now, the only streamline that needs to make a sharp turn at the corner is the streamline that lies exactly on the surface.  Everywhere else, in a subsonic flow, the streamlines can make a gradual turn.
Of course, the limiting streamline can't make an instantaneous turn, so in the subsonic inviscid approximation, we actually have the limiting streamlines go to zero velocity at a sharp trailing edge like this -- that forms the aft stagnation point.
If this airfoil had a cusped trailing edge, things would be a little different.
Of course, all real flows have a boundary layer (even if it is thin), and the limiting streamline at the surface obeys a no-slip condition -- so its velocity is zero.  The rest of the streamlines are able to make the TE turn without a singularity.
So, in our inviscid idealization, I would expect that the normal shock would stay a normal shock until it reached the trailing edge.  At that point, like the limiting streamline in the subsonic case, the  on-body streamline needs to instantly turn and it would do this by going to zero velocity at the trailing edge.
As the freestream Mach is increased further, the shock should lay over and become the oblique shock solution that we expect.
I'm not exactly sure what happens during that transition.  Since this only applies to our idealized model, I'm not sure that those details are relevant at all.  After all:

boundary layers happen.  They allow the limiting streamline to turn and solve the stagnation streamline problem.

real shocks are not infinitely thin

real trailing edges are not infinitely sharp, they have some bluntness to them.

Myriad other complicating factors

I see, thanks for the quick reply. So let me make sure I have this right : The trailing edge shock is the same as the shock on the top of the wing. It just stops at the trailing edge because of the air coming in contact with the air from the lower surface.

No.  The shock on the top of the wing does not cause the flow to turn.  So it is a normal shock, while the TE shock is an oblique shock.  The shocks on the surface move aft and when they get to the TE, they change from normal to oblique.  Does that transition happen just before it reaches the TE, or does it happen at the TE as it speeds up, I'm not sure.  In reality, the airfoil will have a boundary layer that will make some of these distinctions more messy.

Ah I see, that edit cleared some things up. One last little thought I had : If the flow behind the normal shock is subsonic (in sections 2 & 4), how would an oblique shock form where the TE converges to a point? Does the flow have to be supersonic to create an oblique shock?

The flow upstream of a shock must be supersonic.  The flow downstream of a normal shock is always subsonic.  The flow downstream of an oblique shock can be subsonic or supersonic -- depending on the angle of the shock.  There are two solutions, the 'strong' and 'weak' shock solution.  In the second drawing, there would not be a second set of shocks at the trailing edge.

I see, thanks. So small follow up question about the air streams meeting at the TE. The 2 streams (upper and lower surface) will be meeting at an angle at the TE. That means that their vertical velocities will add, which would make them meet at a faster speed. Basically if you run into a wall, but the wall is moving towards you.

When they meet at that point, their velocities must be parallel.  That means that something must be done away from that point to accelerate them (turn them or slow them down).  In a supersonic flow, the shock happens to 'instantly' turn them parallel. In an inviscid subsonic flow, they slow down as they approach that point to reach zero velocity and then accelerate back up away from that point.  Your wall analogy at best applies to the supersonic case -- something abruptly must happen, so you get a shock to turn the flow.  In subsonic flow, abrupt changes are impossible.

Oh okay, that makes sense. Thanks for all of your help on the questions I’ve asked in the past few days, I really appreciate it!


text:
What does the letter in these ATC data blocks indicate?

Why do some have an F and others have a D?

What is the source of the image? What else can you tell us about it (what airspace does it show, etc.)?

@TypeIA It is Sarasota airspace. It is from this Vasaviation video.   https://youtu.be/Wsf6zzmzH8c

Those letters indicate the wake turbulence category of the aircraft.
Based on context—LFA, callsign "BUCKY," is a flight school and is unlikely to by flying "Upper Large" aircraft—the categorization being used in the video is probably the "Wake Turbulence Recategorization" (RECAT) rules, which is an older version of of the wake recategorization project (see FAA JO 7110.659C, the latest version of RECAT in use before its cancellation). The current standard is known as "Consolidated Wake Turbulence" (CWT) and procedures are defined in FAA JO 7110.126. The wake turbulence category of a particular aircraft type can be found in FAA JO 7360.1. If you had taken the screenshot at a slightly different time the datablock would have shown the aircraft type in the righthand side instead of the groundspeed.
Note also that this is NOT a depiction of a real-world FAA system; VASAviation feeds ADS-B data into their own simulator software. It looks pretty close but you cannot assume it is the same thing that the controllers were seeing at the time.

Do you know what the F and the D by the speed indicate?

Yes, and I said that in my answer... "F" indicates RECAT category F and "D" indicates RECAT category D...

Oh sorry, didn't catch that.


text:
What is concentrated payload?
Reading about the Embraer c390 millennium. Two different payloads are listed. Maximum concentrated payload and distributed payload. What is the difference?

A concentrated load is one single very heavy thing that is compact- like a big electric motor strapped down for shipment on a pallet, that takes up little floor space relative to its weight. A distributed load is spread out, like a line of suitcases stretching from the front to the back of the cargo compartment, occupying the entire floor.
Even though that huge electric motor might weigh exactly the same as that long line of suitcases, the stresses imposed by each load on the airframe will be quite different, and they will have distinctly different effects on the pitch response of the plane in flight.


text:
Why does the mixture knob spin?

On my first intro flight. I noticed the instructor twisting the mixture knob. I didn't think of asking, but now Im curious.

Apparently it is designed to be turned, so it probably wasn't just idle twiddling.

What make and model airplane? Maybe we can look it up.

John K has the correct answer below, but this is pretty common on GA planes like Cessnas

Ever see a vise with one of those levers that you hold and then you can quickly pull the vise open a large way in a short time?

It's called a Vernier Control.  It has an internal screw that is part of the fixed housing, with threaded jaws that engage the threads around the outside and are part of the knob.
When you turn the knob, you are turning the jaws with internal threads like a nut, and the whole thing moves in when you turn clockwise (richer) and out when counterclockwise (leaner), moving the teleflex cable running to the carb or fuel injection controller with it.  When you push the center button, it opens the jaws, disengaging the threads and now the knob can be pushed straight in and out like the throttle.
It makes it easy to make very fine adjustments by turning the knob, but when you need to make large changes, you don't have to furiously spin the knob, you just push the center button in with your thumb and you can pull the whole thing in or out.  In an emergency when you need to go "balls to the wall", just shoving the knob forward with the heel of your hand pushes the button in automatically and lets you move it in one motion.
You usually see mixture and propeller controls done this way, because most of the time you are making fine adjustments until those times you need to shove them all the way in or all the way out.  You'll sometimes see Vernier throttles as well, but it's less common.

Probably because the actuation mechanism contains a screw drive which allows fine adjustment of the mixture- either that, or the screw drive locks the pull knob firmly in position so it will not vibrate out of the desired position.

the screw drive locks the pull knob ... so it will not vibrate out of the desired position - This was my first guess too, but why do this for the mixture and not for the throttle?

@A.I.Breveleri IIRC the throttle has a similar mechanism, see the reference to "Throttle Friction Lock" page 4-7 on https://longislandaviators.com/wp-content/uploads/2018/08/1978-Pilots-Operating-Handbook-Cessna-152.pdf

@A.I.Breveleri the mixture knob is locked in place by the screw drive, but can be released for quick adjusment by pressing the center button on the knob, which releases the half nut holding on to the threads. The throttle, on the other hand, has a friction lock, which can be adjusted to find a suitable "stickyness". In the picture, the friction is adjusted by the corrugated aluminium finger screw with "do not lubricate" stanced on it. Lube would obviously spoil the system, so that vibration would make the throttle settin creep.


text:
What is needed in a plane to fly approach mode?
Do you need anything other than approach mode to be able to have the plane fly the ILS to minimums? Do you need a EFD (glass cokpit) or can it be done with steam gages as well?

You’ll need to better specify the equipment available and how it’s connected.  As it is there’s too many unknowns.

What is required for a plane to have that ability?

To fly an ILS in general or to fly a coupled approach (with autopilot)

To be able fly an ILS.

But you accepted the answer specific to flying a coupled approach (with autopilot)?!

In order to "have the plane fly an ILS to minimums" - i.e. an autopilot-coupled approach - you'd need

an ILS receiver (which would be comprised of both localizer & glideslope receivers)
an approach coupler as part of the autopilot
a functional display of the localizer and GS so that the pilot can monitor the autopilot's performance
the necessary approval from the aircraft manufacturer, unless it's an Experimental aircraft
the necessary OpSpec approval in the case of an airline operation
all the other stuff you'd need to fly an ILS by hand (altimeter, etc)

Things you wouldn't need:

glass cockpit (flying a coupled ILS predates glass cockpits by a couple of decades)
digital anything (analog autopilots can do this just fine)

NOTE HOWEVER that a hand-flown ILS approach to minimums is entirely possible in most cases, and none of the stuff related to the autopilot is necessary at all. (Some minima like Cat II or Cat III with certain equipment can be the exception to that "most cases" statement.) And that also pre-dates all digital displays by decades.

Plus the stuff listed in 91.205(d) and for CAT II approaches the stuff in Appendix A of that part.

I’m confused too, why do you need an autopilot?  The question isn’t about coupled approaches.

I read the phrase "to have the plane fly the ILS to minimums" as a reference to a coupled approach, and thus distinct from simply what's required for the pilot to hand-fly the ILS. Re-reading the OP's comments seems to indicate that his intent was otherwise. Though he accepted this answer. I'll edit the answer so it's more clear what my list refers to. @MichaelHall

@RalphJ Thanks!!!

@MichaelHall I just wanted to know what the auto pilot uses to fly an approach.

@Boeing787, well I don't think you did a very good job clarifying that when Jim asked above...


text:
What is this indication on the G1000?

I noticed it came on when glideslope was intercepted. Does anyone know what it is?

That would be the Outer Marker indicator as shown in page 74 of the manual


text:
Have the terms Mayday and Pan Pan become less used?
I've noticed listening to videos of emergencies that pilots simply say that their declaring an emergency vs saying Mayday. Am I wrong?

You're not wrong, but I don't know that it means they are being used less.  Perhaps they were used more in the days of poor radio reception?
I have only declared an emergency twice, but neither time used any phraseology other than "declaring an emergency" as you have heard, and described the issue briefly in plain language.  On clear VHF frequencies when you already have established two way communications there really isn't that all that much reason to say them.  If you are already talking to a controller simply tell them what you need.
I think the real intent of the words is to get everyone else to shut up, but the only time I really needed to do that I instead squawked 7700 and just started doing what I needed to.  (I was VMC with the field in sight on a downwind vector, had already declared minimum fuel after diverting, and was now emergency fuel.  I had been told I was #2 behind a C-130 on a 10 mile straight in who I also had in sight, but there was a lot of chatter and I couldn't break in to express the urgency of my fuel situation.  I turned to a visual base leg, and as soon as my squawk got their attention I was cleared to land immediately.  I don't think the C-130 even had to go around...)
In any situation when you need to make sure you have ATC's full attention the words are certainly one of the best ways to do so, and I wouldn't hesitate to use them depending on the situation.  (poor radio reception, other traffic trying to break in, etc.)  It is still the proper phraseology for declaring an emergency.

I used PAN once, to raise an airliner flying overhead, on 121.5, after I was forced down on a remote lake in NE Ontario from engine failure. I was out of radio range of anybody, didn't want to have to set off the ELT, and I noticed a contrail overhead. The airliner actually responded and contacted my base for me with my location, and me and my two pax (fishermen who happened to be lawyers(!)) were rescued later the same day, by the air service owner, who brought gas cans thinking I was just low on fuel. When I told him I had a blown cylinder on the radio as he approached, he said "Ohh sh*t!!"

@JohnK That's a super cool story. Thanks for sharing!

An American pilot speaking in plain English to an American controller using clear VHF works fine of course, but using standard phraseology is still recommended. Otherwise, not everyone on the frequency might understand what's going on. Chinese pilots are regularly confused by American controllers speaking plain English to them because they only learned standard phraseology. French controllers also love speaking French with French aircraft and nobody else knows what's going on...

As a marine VHF operator, I completely recognise "the real intent of the words is to get everyone else to shut up", particularly as we would normally use the calling channel (16) for such a broadcast message.  Aviation comms is a bit different, particularly if you already have an established channel to ATC, and other traffic is much less likely to be able to stop and help you...

The following is written from the perspective of a guy who hasn't even bothered getting my pilot's license. I've done quite a bit of research on the topic, but I'm hardly the most qualified expert to answer aviation-related questions.
The answer is getting rather long-winded for what started as a casual observation. But the gist of it is that I think we hear "declaring an emergency" from old pilots because it's what they were taught (in America, where much, if not most, aviation content on the internet comes from), and from new, casual pilots in minor emergencies because it sounds, in calm language, less formal (and potentially embarrassing) than "panpan".
Some Background
It should be noted that until a few years ago (2014, if memory serves), "declaring an emergency" was the official phrasing in the United States, while "mayday" or "panpan" were acceptable, but unofficial, alternatives.
The standard phrasing is now "mayday" and "panpan", in keeping with emergency phrasing the rest of the world uses. However, people tend to be creatures of habit, especially when approaching (or past) task saturation limits, so you're likely to continue hearing "declaring an emergency" for several more decades.
Right now, the majority of pilots got their licenses1 more than a decade ago, so the majority of pilots were likely taught the old phrasing. As more old pilots retire and more young pilots get training, the new (to us) phrasing will become more popular.
The U.S. accounts for nearly half of all air traffic in the world, and probably more than half of all English-speaking air traffic in the world. So there's a high chance that whatever flight you're watching, listening to, or reading about is from America, just based on numbers. If you're an English speaker, especially one from the U.S., the odds are especially high.
Alternate Hypothesis
However, I'm going to go with a frame challenge here. You seem to believe "mayday" and "panpan" are becoming less used, but I don't think this is actually true. I would argue the opposite, based on the above information.
I suspect what's really going on is that you watched a lot of movies growing up, where "MAYDAY! MAYDAY! MAYDAY! FOX! SPITFIRE! IN THE HOLE! 72! EVERYONE IS GOING TO DIE!" sounds far more dramatic than "A123XYZ declaring an emergency". Then you became interested in aviation and started looking up real aviation, where "declaring an emergency" has been the standard (at least in this part of the world) for decades.
Alternately, it's possible you're from a part of the world where "mayday" has always been the standard, then you got on YouTube, where much of the content is from America.
Either way, I'd suggest you challenge your belief that "mayday" and "panpan" used to be used a lot, because it's likely either not true, or it's because you're used to a different part of the world.
When Neither "Mayday" nor "Panpan" Make Sense
A "mayday" call is only used when the aircraft is in grave, immediate danger. "Panpan" is used when the aircraft isn't falling out of the sky, but either it's getting close, or there's a danger to someone on the aircraft (a passenger had a heart attack, for example). In either case, you're telling ATC that you need help, right now.
But there are plenty of cases that aren't really dangerous, but you'd really like some kind of priority assistance. You might use a "panpan" call, and I think it would be perfectly legal, but it's a bit extreme. Perhaps the navigation system died, but you're still perfectly good for visual flight in nice, cloud-free weather. You know it's probably not a big deal, but it just gets under your skin. If the nav system died, what other problems do you not know about?
So you calmly tell ATC that you've got some minor problems and you'd like to officially declare an emergency. You explain that you'd like to divert to the nearest airfield with onsite or nearby aircraft mechanics, that you're feeling a little stressed, and you wouldn't mind the biggest runway available just to make getting on the ground as smooth as possible.
Both "mayday" and "panpan" are excessively alarming for a minor situation that's mostly imaginary. But your elevated stress is quite real, and you'd quite really like a bit of a lifeline. It can also help just to have a supportive ear on the other end. So the less alarming phrasing can be used as an informally tertiary alarm call.
With the surge of YouTube channels dedicated to everyday flying, we're seeing more and more of these kinds of "emergency" calls for general aviation being recorded and broadcast, even though the situations themselves aren't really new or unusual.
With large, commercial aircraft, much of the above still holds. They have so many redundant systems it's very rare one falls out of the sky, so it's very unlikely a "mayday" call is warranted. With potentially hundreds of passengers on board, "panpan" calls are more likely to arise, though I can't think of a case I've actually heard that phrase used in a U.S. ATC call. But there are plenty of occasions where a critical enough system fails that the airline (or FAA) requires the airplane to make an emergency landing, but it's still not really a "panpan" situation so much as a formality to avoid getting to those far more dangerous situations.
I'm not sure what the current airlines are teaching, but it seems likely that many of the commercial situations "declare an emergency" precisely because of how safe the aircraft are.
Formal Definitions
The FAA defines an emergency condition as either a "distress" condition or an "urgency" condition. The "mayday" call is used for the former, and "panpan" for the latter. You can see the official procedures involving "mayday" and "panpan" calls in Chapter 6, Section 3 of the AIM.
The official definition of the radio calls can be found in the AIM's Glossary for MAYDAY

MAYDAY- The international radiotelephony distress signal. When repeated three times, it indicates imminent and grave danger and that immediate assistance is requested.

and for PAN-PAN.

PAN-PAN- The international radio-telephony urgency signal. When repeated three times, indicates uncertainty or alert followed by the nature of the urgency.

Chapter 6, Section 1, Paragraph 2 of the AIM says (emphasis mine):

An emergency can be either a distress or urgency condition as defined in the Pilot/Controller Glossary. Pilots do not hesitate to declare an emergency when they are faced with distress conditions such as fire, mechanical failure, or structural damage. However, some are reluctant to report an urgency condition when they encounter situations which may not be immediately perilous, but are potentially catastrophic. An aircraft is in at least an urgency condition the moment the pilot becomes doubtful about position, fuel endurance, weather, or any other condition that could adversely affect flight safety. This is the time to ask for help, not after the situation has developed into a distress condition.

Pilots who become apprehensive for their safety for any reason should request assistance immediately. Ready and willing help is available in the form of radio, radar, direction finding stations and other aircraft. Delay has caused accidents and cost lives. Safety is not a luxury! Take action!

Clearly, the FAA isn't worked up about inappropriate calls of "panpan". "The moment the pilot becomes doubtful" is a very low threshold to meet. And the definition of the call is simply "uncertainty or alert".
Still, it's a formal command that feels more forceful (at least to me) than a more casual, but still explicit "declaring an (unspecified level of) emergency". To me, and seemingly to many pilots, "panpan" feels better reserved for a situation where immediate assistance is needed, even if that's not the formal definition.
I'm in no way saying "panpan" shouldn't be used. Just that many pilots seem hesitant to use it, and I think the reason is that "declaring an emergency" in a casual tone feels like a less embarrassing way to say "I'm out of my depth" than the more formal calls.
I've also heard "I'MDECLARINGANEMERGENCY! I'MDECLARINGANEMERGENCY! I'MDECLARINGANEMERGENCY!" used a minute or two before a guy crash-landed on a freeway offramp. The tone and speed of the declaration made it very clear it was a distress call, and that "MAYDAY! MAYDAY! MAYDAY!" is both easier and faster to say. But it also illustrates the versatility of language via intonation, and how humans perceive the same statements differently depending on context.
1 Now called "certificates", though the FAA still uses the term "license" in a lot of their literature, last I checked.

Just stating that you are "declaring an emergency" and prefixing your call with "Mayday Mayday Mayday" have exactly the same meaning. Saying that you are "declaring an emergency" is more severe than "Pan-Pan," not less. If you have an urgent (or distress) situation, there is no need to be shy about declaring. No one is going to tell you your emergency wasn't emergent enough for a Mayday call. If you get more attention and priority handling than was strictly necessary, no one is going to die. Lots of people have died because of being too hesitant to declare when they should have.

@reirab: At this point, an "emergency" is either a distress (mayday) condition OR an urgency (panpan) condition. Declaring "an emergency" could be either condition. You can see the FAA definitions in play here, as well as other sections of chapter 6. I agree (and the FAA stresses) that calling "panpan" for something that turns out to be nothing is legal, but pilots still get stressed over it. Legally, "declaring an emergency" has the same impact as "panpan" or "mayday", but I think the apparent stress level is

...higher when you call "panpan" (a formal emergency call) than stating something like "I'm going to go ahead and declare an emergency". Of course, you could also state "I'm going to go ahead and declare panpan". But that (to me) sounds weird, and the point of the "mayday" and "panpan" calls is specifically so the individual words gets other radio traffic to stop. Why use the forceful term when you've already got the air?

Why about "sécurité"?

@Someone: As far as I can tell, "sécurité" is used in maritime emergency calls (see here for example), but not aviation in America. The word is too common in French for me to easily find examples of it being used in foreign aviation, if at all.

@Someone "sécurité" isn't used in aviation (at least not in any internationally-standard manner... of course, French speakers would use it as normal French when communicating with each other in French, but not with the standard meaning it has in maritime radiotelephony.) The word does not appear at all in the ICAO Manual of Radiotelephony or the FAA Pilot/Controller Glossary, for example. Which is probably for the best, as it would be confusing with the English word 'security,' which is used in the P/CG and does not mean the same thing.


text:
How is the pressure constant across the boundary layer but density isn't?
In $P=ρRT$, ρ is density, R is the gas constant, and T is the temperature in Kelvin. In the boundary layer, the temperature and density varies, so why doesn't the pressure?
If you have a slip line in a shock system, the density might be different across it and the pressure might be the same. This is different than the pressure/density/temperature relation in the BL, because the air went through different shock systems in the slip line example, so the properties are bound to be different.
Basically, in this specific example, I'm not sure how increased temperature/density wouldn't increase the pressure.

downvoter, please explain why you downvoted. I can't improve anything unless I know what's wrong with this question...

Because the pressure is constant across the boundary layer.
If the pressure at the wall was higher, it would do something strange -- such as push the streamlines away from the wall.
The pressure at the wall of a boundary layer is set by the pressure on the outside of the boundary layer.  Period.  Full Stop.  You don't get to set it twice.
Increasing temperature does not increase pressure.
The fact that $P$ is on the left in $P=\rho R T$ gives it no special place in the world.  We just prefer multiplication to division, we can equally well write $T=\frac{P}{\rho R}$ or $\rho=\frac{P}{T R}$.
You might be thinking about Boyle's law, Charles's law, or other such things.  These are each special cases of the $P=\rho R T$ gas law.  They are limited to constant volume, constant pressure, or constant something else.
Edits:
There are lots of situations we can think of where pressure and volume will be related.  Here are four containers of gas....

The first is totally sealed and rigid.
It is a constant volume container.  Since it is sealed, it is also a constant density container.  If you heat it up, the pressure will change because the density is constrained to be constant.
The second has an idealized piston.  We imagine that this piston creates a perfect seal, but has no friction and no mass.
This is a constant pressure container.  The pressure inside the container will always be equal to the ambient pressure outside the container.  The amount of gas inside the container (mass) is constant, but if you (for example) cool the gas, it will shrink and the piston will move down to maintain constant pressure.
The third has an idealized piston, but there are a stack of weights on top of the piston.  It also forms a constant pressure volume, but this time the pressure is the sum of the ambient pressure and also the amount of weight divided by the area of the piston.
The fourth is a fun one.  On top of our idealized piston, there is a spring that is rigidly fixed at the top.  What do you think the pressure-volume relationship will be for this case?
These are simple examples of physical systems where it is easy to see and understand the relationship between pressure and volume (and hopefully thereby density and pressure).
In all of these cases, $P=\rho R T$.
In each of these cases, something else determines the relationship between pressure and volume (or whatever the process entails.)  Here, that is determined by the structural rigidity of the box, by the behavior of the piston, by the stack of masses, or the deflection of the spring.  I.e. additional physical principles and equations are needed to specify the PV relationship.
In fluid dynamics, the additional physical principles that determine the PV or P$\rho$ relationship are the conservation of mass, momentum, and energy.
BTW, The best (and easiest) way to render $\rho$ here is \$\rho\$.

Oh okay, thanks for your answer. If you do an example equation for P = ρRT, you get a certain number for P. If you increase T, in my case, I got a higher number for P. Does that not apply for the BL? Or does it not apply for the BL because it's not a confined space?

@Wyatt As long as the flow is free to expand, all a temperature increase does is to lower density, so more volume is taken up by the same amount of air. P will only increase if volume is constrained (such as inside a pressure cooker).

The boundary layer is not a confined space.  Instead, it is a constant pressure space.


text:
Wouldn't the induced downwash "pull" the wing of a plane down?
I asked a similar question on this recently, (here), and the answer said this:

Downwash is not a force. It is a small increment in the velocity vector.
Downwash can change the local angle of attack -- which can change the lift -- which is a force.
However, A wing has a rolling moment because the force on the two sides is not equal. Period. Full Stop. If the lift force on the left wing is greater than on the right, the wing will roll to the right. It is that simple.
We can talk about where the force imbalance comes from -- but you won't have a moment without a force imbalance.

This makes sense, but if the downwash changes the direction of the air above the wing, wouldn't that in a way push on the upper surface of the wing?
As seen here, the black arrow represents the velocity vector imparted on the flow because of the induced downwash (from tip vortices). Wouldn't that air then go on to 'hit' the wing, pushing it down? The answer above says downwash isn't a force, but wouldn't this count as a force from downwash?
(apologies for the bad drawing of the arrow)

Related: https://aviation.stackexchange.com/q/50301/34686

NO.
First, this is a 2D image and represents a 2D flow.  There are no tip vortices (trailing vortices) in a 2D flow.  Instead, it is all about the bound vortex (at the quarter chord).
Notice the upwash in front of the airfoil.  This is also because of the bound vortex.
The act of creating lift causes the streamline in front to come up -- and the streamlines behind to go down.
The upwash / downwash does not cause the force.

I think I understand what you're saying. So the downward bending streamlines on the rear side of the airfoil isn't from the induced downwash, which makes sense. But would the induced downwash (on a 3d wing where there are tip vortices) create an extra 'bending' of those streamlines downward, which would intersect with the wing and push it down? See this photo to maybe see what I mean. (Sorry if I'm not picking up on something obvious)

You can't separate these effects the way you're trying.  It all goes together.
The downwash after the airfoil or wing is downwash, it is caused by the presence of the wing, it influences the flow on the wing.  However, the solution (i.e. the forces and flowfield you see) is a coupled combination of these effects.   You can't separate them out artificially.  Everything effects everything -- and that has already been accounted for.

Again, I emplore you.  Stop using crappy internet resources.  Read the engineering textbooks you have.

Yeah that’s probably a good idea; anyone can post anything on the internet. Even if someone is trying to post useful resources, they themselves might be misinformed.

It might help for you to draw an arrow up over the wing and down after the wing to represent the balance of forces involved with lift.
It is intuitively correct that the arrow hitting the top of the wing is bad.  This can happen when angle of attack exceeds the critical stall limit.
It is important to understand the wing must have sufficient forward velocity to pass by a given parcel of air before the downwash hits the back of it, collapsing the low pressure area that creates lift.
Key to understanding this is the concept that air molecules have mass, therefor inertia. Get them moving up and a downward force can only return them to their original position in a given amount of time.  By that time the wing has passed.
Now on to wing tip vorticies.  There one would try to design a wing tip to keep the vortex as far away from the lift producing area of the wing as possible, so it's downward rotation cannot strike the top of the wing. Hoerner wingtips push the outflow from the bottom of the wing further away from the wingtip.  Swept wings can use winglets to both minimize and move the vortex further away.  This increases the lifting efficiency of the wing, allowing for fuel saving lowering of AoA to create the same amount of lift.


text:
Why does a lower chord length reduce the pitching moment?
I've heard that higher aspect ratio wings (also lower chord length wings) are more stable and less maneuverable. Why is that?

The question in the title is different than the one in the text. Is your doubt more about wing pitching moment or airplane stability?

@sophit I kind of had 2 questions. Doesn’t the wings pitching moment relate to stability and maneuverability?

Wing pitching moment has no direct influence on stability and manoeuvrability is a quite broad terminology where wing pitching moment might or might not have an influence

@sophit ah okay thanks for the clarification

You're welcome. Some time ago I elaborated an answer about stability in pitch but I definitely need to streamline it a bit. I let you know when/if I'm done

Oh alright thanks for that!

Done, let me know if it is understandable

Let's translate some of this to layman's terms

higher aspect wings contribute less to stability

The portion of the wing aft of the CG contributes to stability.  It's torque consists of area × distance from CG.  A wing with a longer chord will be more stable.

higher aspect wings are less maneuverable

Generally not because of higher stability, but because of lower G load tolerance (its easier to break a long thin stick compare with a short thick one).
Those who endeavor to increase aircraft fuel economy by increasing aspect ratio (airliners) should realize a larger, not smaller, tail may be needed to maintain the same stability characteristics as the previous iteration.

Ah okay so high aspect wings are generally not more stable in pitch naturally? Edit : just re read your answer and saw the part about the section of wing behind the CG. Thanks!

@Wyatt you can check Cm data at airfoiltools.com.  Profile of wing (degree of camber) can also affect stability, as well as (as you have noted) CG placement.

One question though : does that mean that high chord length wings are more or less maneuverable than low chord wings? (Saw the part about more chord behind the cg but I wasn’t sure about maneuverability) edit : didn’t see your previous response until I commented this

@Wyatt you can make something more maneuverable by increasing the area of the control surface.  Remember the distance from CG is far greater at the tail than at the wing trailing edge.  Strength (G load limit) is the limiting factor, not chord length.  Look at an F22 Raptor.

Oh okay, makes sense. A question I had was say you had most of the wing forward of the CG (unstable), would a larger chord wing still be more stable than a smaller chord wing in that instance?

@Wyatt that will depend on the wing profile.  The wind tunnel is your friend here, but applications of rear CG placement are rare.

Ahh, okay makes sense. Thanks a lot!

Just as we non-dimensionalize wing lift coefficient as
$C_L=\frac{L}{q\,S}$
We non-dimensionalize wing pitching coefficient as
$C_M=\frac{M}{q\,S\,c}$
Where the chord provides the additional length unit to get a force to a moment.  However, this also shows you how the forces and moments scale.
The moment on a wing is proportional to the pitching moment coefficient, ($C_M$), dynamic pressure ($q$), wing reference area ($S$), and wing reference chord ($c$).
$M=C_M\,q\,S\,c$
So, if you have two wings that are otherwise the same, then the one with the larger chord produces the larger moment.
The other thing to consider is that a wing does not act alone -- you should also consider the horizontal tail when considering an aircraft.
We usually non-dimensionalize the size of a horizontal tail using the horizontal tail volume coefficient...
$V_h=\frac{S_h\,l_h}{S_w\,c_w}$
Where the $h$ subscripts are for the horizontal tail and the $w$ subscripts are for the wing.
I.e. a horizontal tail's effectiveness is proportional to how big it is ($S_h$) and how far away it is ($l_h$ the tail moment arm).  And inversely proportional to the wing's size and chord.
The wing terms are the terms that non-dimensionalize the wing's moment coefficient.
If you think about a horizontal tail's contribution to pitching moment as mostly due to lift over a moment arm, you will see that the $S_h$ comes from the lift of the tail and the $l_h$ is the tail moment arm.

thanks for your answer. One question that I'm not sure if you answered in you response : Does the chord length affect the pitch maneuverability of the plane? (you might've answered that in your response already but I didn't go to aeronautical engineering school so I didn't understand everything) Thanks!

Yes and no. We usually think about stability in terms of the static margin - as a fraction of the reference chord. So a 5% SM aircraft is much less stable than a 20% SM aircraft. However, if we dimensionalize those values to say 1ft and 4ft, then we think about changing the chord of the aircraft, we see that a smaller chord will make the same dimensional SM be a larger %SM. Consider a cargo aircraft that needs to tolerate changes in the center of gravity. Those would be scaled by the chord -- such that a small chord aircraft would be less tolerant of CG movement than a large chord one.

However, if we consider maneuverability via the tail volume coefficient -- which is a simple measure of the tail effectiveness.  We see that cutting the wing chord in half (keeping everything else the same) would double the tail volume, making for a more sensitive and maneuverable aircraft.

thanks for that info. If the SM is the distance of the CM from the NP, how does that affect stability? I thought that the main judge of stability was the distance of the CP from the CM. Also, could you put the info you provided me in laymans terms, I don't know some of the terms used, and other things like that. Thanks again.

(forgot to notify, didn't know if you'd see my comment if I didn't, haven't used stack exchange for that long so I don't know much)

@Wyatt The CP changes with angle of attack, so it is not a useful point in terms of stability.  The NP does not change with angle of attack.  Your intuition about distance from CM applies to the NP.  When an aircraft is trimmed (at some alpha), the elevator has been adjusted such that the CP is at the CM.

oh okay, makes sense. I saw this on a different question : “ The lower chord also reduces the wing's pitching moment, so a smaller tailplane can be employed, or the tail lever arm of the horizontal tail can be shortened.” Does that mean basically what you were saying, as the same dimensional SM is a larger % SM on a smaller chord wing? Also if a larger chord wing produces a larger moment, how do 2 planes that are the exact same, but the only difference being the smaller chord length of one, how are they more maneuverable (the one with smaller chord) like you said in a previous response.


text:
How do 2 vortices (or more) combine when they are near each other?
If you have 2 vortices rotating the same direction, how will they interact with each other? For example, if you have 2 tip vortex sized vortices close to each other and rotating the same direction, how will they combine? Will they have any sort of 'suction' or attracting force towards each other? If so, why is that?
(I would ask what would happen if they were counter-rotating, but this question already has answers)
Say you have 2 wheels that are spinning in the same direction. if you try to put them together like gears, they will try to cancel each other out, or try to spin the other one the opposite direction. That's why I'm confused as to how vortices combine.
Thanks, and happy new years!

Intuition is tricky because vortex merging depends on many things.  Be perplexed by these two paper abstracts, for example.

You can linearly combine the speeds caused by vortices. This can create constructive or destructive interference; this means the speeds induced by either one can either add up or cancel with those of the other. If you have two vortices rotating the same direction next to each other, they will combine into a single, stronger vortex.

oh okay, thanks. So is my analogy of the 2 wheels spinning in the same direction incorrect? It seems like the same principle, so it's confusing as to why they add to make a bigger vortex.

The analogy is correct.  Where two clockwise wheels meet, they cancel out "inside" of the bigger combined vortex.  But their outsides survive, like a chain or belt around them both.

@CamilleGoudeseune ah okay, thanks. Yeah I kind of thought something like that would happen. One last question, do 2 vortices next to each other have any attracting force?

@Wyatt as the answer says, the vortices will combine so you could say that they attract each other, but remember that a vortex isn’t a ‘thing’ in its own right, it’s differential movement of air.

I just developed a small question about this, and I thought this would be the right place to ask. If you have 2 vortices rotating the same direction, why would they make a stronger one once combined? Like explained in CamilleGoudeseune's comment, the first half of the 2 vortices will be canceled out. Then you're left with a half of 2 vortices which would combine to make the same size vortex you started with? Like if you cut 2 wheels in half and then took the remaining outer half's and put them together. If that doesn't make sense I can try to explain it better. Thanks

@Wyatt No, the resultig vortex is the sum of both. When the adjacent halves cancel each other, what remains are two outer halves at a wider distance. Since vortex velocity is higest at the center and drops inversely with distance from the center, the two halves at a wider distance combine into a stronger vortex.


text:
How does a shear layer make vortices?
How does a shear layer make vortices? If you have 2 streams of air that are moving at different speeds or opposite directions right beside each other, how would that make a vortex? What causes the rotation, in other words.

A shear layer is no different than a boundary layer.  Except both sides are a fluid vs. fluid/solid.  And usually we choose a reference frame that has the solid frame at zero velocity -- and frequently, both sides of a shear layer have non-zero velocity (but we could choose a frame such that one side is zero and it would be the same flow).
Put your hands together flat like you are clapping.  Move your hands -- they can only move in the same direction (or they pass through each other, or you have a void between them, which we disallow).  If they move at different velocities, there is shear (friction) between them.  The fast one tries to speed up the slow one - and the slow one tries to slow down the fast one.  However, if they move the same speed, there is no friction.
Think about an element (tiny square) of fluid.  Pressure acts perpendicular to the faces of the square.  Shear acts parallel to the faces of the square.
Shear force is proportional to a change in velocity -- if velocity of two streamlines is the same, there is no shear between them.  No viscous force.
Shear in a fluid causes vorticity.
You should check out these videos.  In particular, watch this one.

oh okay, thanks for your help! So as you said, shear force acts parallel. If it acts parallel, what force is causing the rotation/curve needed for a vortex? Sorry if I'm not picking up on something obvious. Also, interesting video! I find it interesting how such an old video can still be straight forward and helpful.

The Shapiro videos are great.  Some of the best discussions of vorticity and other phenomena.  There are two on Vorticity, (I linked to the first), watch them both.
The shear forces cause the fluid element to deform.  Rotation is a part of that deformation.   This is similar to solid mechanics and should be covered in any textbook.

Ah okay. I watched the first one, and if you look at time 12:50, you'll see a little sketch he did. Is the sideways x denoting the flow encountering a disturbance and turning into a vortex?  It goes from a + to a sideways x, then to a vortex. I assumed the sideways x was when the flow encountered something which then caused a rotation.

I believe the X is what happens to the flow marked + when it encounters a vortex.

Oh okay, thanks. I think I understand more now that I've thought of it for a while. So I'll just say what I think it is. The faster flow trying to speed the slower one up causes a friction between the streams. That means the faster flow pulls the slower flow, and in doing so makes a vortex. Of course not in flow that is too turbulent or anything.

That is pretty much it.

Ah okay, thanks for your help! I think for me, thinking of the air as a bunch of little balls helps you see why there would be rotation. (Of course that doesn't always work for everything)


text:
What is the purpose of WAAS?
Does it just help with all of the approaches like LPV and stuff? Or is it its own thing?

Have you read through the FAA's WAAS description and benefits?

Global Navigation Satellite Systems (GNSS), e.g. GPS, have a statistical accuracy of about 100m, and it’s particularly bad in the vertical dimension. This imposed severe limitations on how close older RNAV approaches (now called LNAV) could let you get to terrain and obstacles.
Satellite-Based Augmentation Systems (SBAS), e.g. WAAS, improve that accuracy to about 10m. Less position error, especially in the vertical dimension, means you can safely be much closer to terrain and obstacles, which led to new classes of RNAV approaches with much lower minima, e.g. LPV.
SBAS also counts as an extra GNSS satellite for Receiver Autonomous Integrity Monitoring (RAIM) purposes, so an SBAS-equipped aircraft may be able to fly in places/times where a non-augmented GNSS aircraft can’t see enough satellites.

So you're saying it could get you lower minimums?

WAAS includes a ground component (GBAS) providing corrections to GPS signals for atmospheric delays due to, e.g., expansion of the ionosphere from solar activity.

@FosterBoondoggle GBAS is LAAS. Do not confuse the WAAS ground stations for GBAS.

SBAS improves Accuracy and Availability, indeed, but the game changer for aviation is the introduction of Integrity, and enabling integrity monitoring for aviation receivers. That is key for the approval of automated landing based on GNSS.

@FosterBoondoggle do you have a source that the ground component of WAAS is called GBAS?  I have never seen it called that for WAAS.  I always thought GBAS was specifically regarding the GLS ground component, not the WAAS ground component.

Regarding "older RNAV approaches (now called LNAV)" -> isn't the approach still called RNAV? LNAV is just one of the three types of minima (others being LNAV/VNAV and LPV), right?

Too bad the technology lags so far behind commercial and consumer technology.  Thanks to an antenna developed several years ago (at U of Texas IIRC) that allowed GPS receivers in cell phones to locate position in 3 dimensions to within centimeters, which has made the whole drone delivery concept feasible.  I can only imagine the time and cost to certify that kind of capability for an aviation system.

@JohnK In practice, the accuracy of even non-augmented GPS is far better than the statistical models predict. But that’s the number ICAO et al insist on using for manned aircraft.  It comes down to the various errors usually canceling out, whereas the models assume the worst case of additive errors.

@StephenS & Ryan M - I thought I had a source but I can't find it & maybe I misread. Per the FAA you are right that it's specific to local landing system enhancement. "A Ground Based Augmentation System (GBAS) augments the existing Global Positioning System (GPS) used in U.S. airspace by providing corrections to aircraft in the vicinity of an airport..." My comment was meant to observe that WAAS has ground component, which the reply to the OP implied was purely satellite based. Thanks for the correction.

@FosterBoondoggle: "WAAS includes a ground component (GBAS) providing corrections to GPS signals", no GBAS and SBAS are two systems using permanent ground GNSS receivers to compute GNSS corrections and check GNSS integrity. The difference is a SBAS (e.g. WAAS or EGNOS) broadcasts corrections and integrity data via GEO satellites, and a GBAS via VHF ground stations. In both cases, corrections and integrity data are determined by comparing the GNSS positions with the known positions of the receiving stations.